Formulae
1. Alligation:
It is the rule that enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of desired price.
2. Mean Price:
The cost of a unit quantity of the mixture is called the mean price.
3. Rule of Alligation:
If two ingredients are mixed, then
\( {{Quantity of cheaper} \over {Quantity of dearer}} \) = \( {{C.P. of dearer - Mean Price} \over {Mean price - C.P. of cheaper}} \)
We present as under:
∴(Cheaper quantity) : (Dearer quantity) = (d - m) : (m - c).
4. Suppose a container contains x of liquid from which y units are taken out and replaced by water.
After n operations, the quantity of pure liquid = \( [x {({1 - {y \over x}})^n}] \) units.
Exercise : Alligation or Mixture MCQ Questions and Answers
Question 1
A container contains 40 litres of milk.From this container 4 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container.
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| B. |
| C. |
| D. |
Correct Answer : D. 29.16 litres
Description :
Amount of milk left after 3 operations
\( {(40(1 − {4 \over 40})^3)} = (40 \times {9 \over 10} \times {9 \over 10} \times {9 \over 10}) \) = 29.16 litres
View Answer
50/50
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Question 2
An alloy contains zinc, copper and tin in the ratio 2 : 3 : 1 and another contains copper, tin and lead in the ratio 5 : 4 : 3. If equal weights of both alloys are melted together to form a third alloy, then the weight of lead per kg in new alloy will be:
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| B. |
| C. |
| D. |
Correct Answer : C. \( {1 \over 8} \) kg
Description :
Ratio of Zinc, Copper and Tin is given as,
Z : C : T = 2 : 3 : 1.
Now, let the first alloy be 12 kg (taken as 4 kg Zinc, 6 kg Copper and 2 Kg Lead). Weight of second alloy = 12 kg as, C : T : L = 5 : 4 : 3. (taken as 5 kg Copper, 4 kg Tin and 3 Kg Lead.)
Alloys are mixed together to form third alloy. Then the ratio of content in it,
Z : C : T : L = 4 : (6 + 5) : (2 + 4) : 3
Weight of third alloy = 12 + 12 = 24 Kg.
So, weight of the Lead = \( {3 \over 24} \) = \( {1 \over 8} \) kg.
Question 3
There are two vessels A and B. Vessel A is containing 60L of Pure milk and vessel B is containing 28L of pure water. From vessel A, 8L of milk is taken out and poured into vessel B. Then, 6L of mixture (milk and water) is taken out from vessel B and poured into vessel A. What is the ratio of the quantity milk in vessel A to the quantity of pure water in vessel B?
| A. |
| B. |
| C. |
| D. |
Correct Answer : A. 16 ∶ 7
Description :
GIVEN :
There are two vessels A and B. Vessel A is containing 60L of Pure milk and vessel B is containing 28L of pure water.
From vessel A, 8L of milk is taken out and poured into vessel B. Then, 6L of mixture (milk and water) is taken out from vessel B and poured into vessel A.
CALCULATION :
Initial quantity of milk in vessel A = 60L
Initial quantity of water in vessel B = 28L
Now, 8L milk of is poured to vessel B
Quantity of milk left in vessel A = 60 - 8 = 52 L
Quantity of milk in vessel B = 8L
Quantity of water in vessel B = 28L
(Milk ∶ water) in vessel B = 8 ∶ 28 = 2 ∶ 7
Now, 6L of mixture (milk and water) is taken out from vessel B and poured into vessel A.
Quantity of milk in 6L = 6 × 2/9 = 4/3
Quantity of water in 6L = 6 × 7/9 = 14/3
Now, Quantity of milk in vessel A = 52 + 4/3 = 160/3
Quantity of water in vessel B = 28 - 14/3 = 70/3
∴ Required Ratio = 160/3 ∶ 70/3 = 16 ∶ 7
Question 4
Two solutions S1 and S2 contain whisky and soda in the ratio 2 : 5 and 6 : 7 respectively. In what ratio these solutions be mixed to get a new solution S3, containing whisky and soda in the ratio 5 : 8 ?
| A. |
| B. |
| C. |
| D. |
Correct Answer : B. 7:9
Description :
Let the amount taken from S1 be 7x
And amount taken from S2 be 13y
(2x + 6y)/(5x + 7y) = 5/8
16x + 48y = 25x + 35y
9x = 13y
x/y = 13/9
Actual ratios of amounts
= 7x/13y
= (7/13) * (13/9)
= 7/9
Question 5
A mixture is made by mixing alcohol and water in the ratio 9 ∶ 7. If ‘x’ litres of alcohol and ‘3x’ litres of water is mixed in 80 litres of mixture, the new ratio becomes 13 ∶ 14. Find the quantity of new mixture.
| A. |
| B. |
| C. |
| D. |
Correct Answer : D. 108 litres
Description :
Sum of first ratio = 9 + 7 = 16
Quantity of alcohol in 80 litres of mixture = 9/16 × 80 = 45 litres
Quantity of water in 80 litres of mixture = 7/16 × 80 = 35 litres
When ‘x’ litres of alcohol and ‘3x’ litres of water is added,
Quantity of alcohol in new mixture = 45 + x
Quantity of water in new mixture = 35 + 3x
But, ratio of new mixture = 13 ∶ 14
⇒ (45 + x)/(35 + 3x) = 13/14
⇒ 14(45 + x) = 13(35 + 3x)
⇒ 630 + 14x = 455 + 39x
⇒ 39x – 14x = 630 – 455
⇒ x = 175/25 = 7
∴ Total quantity of new mixture = 45 + x + 35 + 3x = 80 + 4(7) = 80 + 28 = 108 litres
Question 6
A milkman claims to sell milk at its cost price, still, he is making a profit of 30% since he has mixed some amount of water in the milk. What is the % of milk in the mixture?
| A. |
| B. |
| C. |
| D. |
Correct Answer : A. 76.92 %
Description :
Let the milk he bought is 1000 ml
Let C.P of 1000 ml is Rs. 100
Here let he is mixing K ml of water
He is getting 30% profit
=> Now, the selling price is also Rs. 100 for 1000 ml
=> 100 : K%
= 100 : 30
10 : 3 is ratio of milk to water
=> Percentage of milk = 10 x 100/13 = 1000/13 = 76.92 %
Question 7
The ratio of Iron and Zinc in an alloy is 4 : 5. In another alloy, the ratio of Iron, Copper and Zinc is 3 : 2 : 7. Equal amounts of the two alloys are molten and mixed together. What will be the ratio of Iron, Copper and Zinc in the resultant alloy?
| A. |
| B. |
| C. |
| D. |
Correct Answer : C. 25 : 6 : 41
Description :
In first alloy, ratio of iron and zinc is 4 : 5
In second alloy, ratio of iron, copper and zinc is 3 : 2 : 7
Suppose, amount T of both alloys is taken.
⇒ Amount of iron in first alloy = (4/ (4 + 5)) × T = (4/9) T
Amount of zinc in first alloy = (5/ (4 + 5)) × T = (5/9) T
Amount of iron in Second alloy = (3/ (3 + 2 + 7)) × T = (1/4) T
Amount of Copper in second alloy = (2/ (3 + 2 + 7)) × T = (1/6) T
Amount of zinc in second alloy = (7/ (3 + 2 + 7)) × T = (7/12) T
After mixing
Total amount of Iron = (4/9) T + (1/4) T = (25/36) T
Total amount of copper = (1/6) T = (6/36) T
Total amount of zinc = (5/9) T + (7/12) T = (41/36) T
⇒ Ratio of Iron, Copper and Zinc = (25/36) T : (6/36) T : (41/36) T
= 25 : 6 : 41
Question 8
A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?
| A. |
| B. |
| C. |
| D. |
Correct Answer : C. \( 1 \over 5 \)
Description :
Suppose the vessel initially contains 8 litres of liquid.
Let x litres of this liquid be replaced with water.
Quantity of water in new mixture = (3 - \( { 3x \over 8 } \) + x) litres
Quantity of syrup in new mixture = (5 - \( 5x \over 8 \)) litres
∴ ( 3 - \( 3x \over 8 \) +8 ) = (5 - \( 5x \over 8 \) )
⇒ 5x + 24 = 40 - 5x
⇒ 10x = 16
⇒ x = \( 8 \over 5 \)
So, part of the mixture replaced = \( {8 \over 5} x {1 \over 8} \) = \( 1 \over 5 \).
Question 9
8 litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of the water is 16 : 65. How much wine the cask hold originally?
| A. |
| B. |
| C. |
| D. |
Correct Answer : C. 24
Description :
Let the quantity of the wine in the cask originally be x litres
Then, quantity of wine left in cask after 4 operations
= \( {{[x(1-{8 \over x})^4}]} \) litres
∴ \( [{{x(1-{8 \over x})^4} \over x}] \) = \( 16 \over 81\)
⇒ \( [ {1 - {8 \over x}} ]^4 \) = \( ({2 \over 3})^4 \)
⇒ x =24
Question 10
How much coffee of variety A, costing Rs. 5 a kg should be added to 20 kg of Type B coffee at Rs. 12 a kg so that the cost of the two coffee variety mixture be worth Rs. 7 a kg?
| A. |
| B. |
| C. |
| D. |
Correct Answer : C. 50 kg
Description :
As per the rule of alligation,
Quantity of Dearer: Quantity of Cheaper = (12-7) : (7-5) = 5:2
Quantity of Variety A coffee that needs to be mixed ⇒ 5:2 = x:20
⇒ x =50 kg